设随机变量$ X $的条件密度函数来自如下形式的单参数指数族

$ \begin{eqnarray} f(x|\theta) = C(\theta)h(x)\exp{\{\theta T(x)\}}, \; \; x\in(a, b) \end{eqnarray} $

(1.1)

其中$ -\infty\leq a<b \leq +\infty $, $ h(x)>0 $, $ C(\theta) = \frac{1}{ \int^{b}_{a}\exp{[\theta T(x)]}h(x)\mathrm{d}x} $, $ \theta $为未知参数, 参数空间为$ \Theta = \big\{\theta;0< \int^{b}_{a}\exp{\{\theta T(x)\}}h(x)\mathrm{d}x<+\infty\big\} $, 利用Holder不等式易知$ \Theta $为一区间. 单参数指数族包括许多重要的分布, 比如指数分布、给定一参数时的正态分布、Gamma分布、Beta分布、Pareto分布、Burr XII分布等, 因此, 研究此分布族的参数估计是有意义的.

视$ \theta $为随机变量, (1.1)为给定$ \theta $时的条件密度函数. 设先验分布$ G(\theta) $为$ \Theta $上的分布函数, 但是无法确定$ G(\theta) $的具体形式, 故此时无法利用通常的贝叶斯方法估计$ \theta $, 因而我们考虑所谓的经验贝叶斯估计.

已有众多学者利用经验贝叶斯方法来对形如(1.1)的单参数指数族的参数进行估计或检验. 在$ T(x) = x $的情况下: (1979)Singh^[1]研究了独立样本时该单参数指数族参数的经验贝叶斯估计, 并讨论了其收敛速度, (2006)陈玲等^[2]与(2008)王惠亚等^[3]分别在NA样本与PA样本下对该单参数指数族参数的经验贝叶斯估计的收敛速度及进行了讨论, (2016)雷庆祝等^[4]在样本为平稳且强混合情形下研究了该分布族的经验贝叶斯估计的收敛速度. 在$ T(x) = -x $的情况下: (2007)陈家清等^[5]与(2008)严拴航^[6]分别在NA与PA样本下利用不同的损失函数分别给出了此分布族经验贝叶斯检验函数的收敛速度. (2013)彭家龙等^[7]利用递归核估计构造了Burr XII分布的经验贝叶斯估计, 并在独立样本下给出了收敛速度.

以上论文均基于一条历史样本$ \{X_{1}, \cdots, X_{n}\} $来构造经验贝叶斯估计, 但在某些情况下(比如相同环境下进行的$ m $组同种实验), 可能会出现多条平行的历史样本$ \{X^{(k)}_{1}, \cdots, X^{(k)}_{n}\} $, $ k = 1, \cdots, m $, 此时, 可以考虑利用平均化的思想来构造经验贝叶斯估计. 本文即利用平均化思想, 在出现平行历史样本的情况下, 结合递归核估计来构造经验贝叶斯估计, 同时, 在强混合样本的情况下得到了该经验贝叶斯估计的收敛速度, 并在强混合的特例: $ \beta $混合下对该经验贝叶斯估计与贝叶斯估计的风险差进行模拟.

设历史样本$ \{(X_{i}, \theta_{i}), 1\leq i\leq n\} $来自总体$ (X, \theta) $, $ X_{i} $是可观测样本, $ \theta_{i} $不可观测. 令$ f(x) = \int_{\Theta}f(x|\theta)\mathrm{d}G(\theta) $为$ X $的边缘分布, 其中$ G(\theta) $为$ \theta $的先验分布(未知). 首先考虑单个样本下参数$ \theta $的贝叶斯估计, 损失函数为平方损失$ L(\theta, a) = (\theta-a)^{2} $时, 参数$ \theta $的贝叶斯决策为后验均值, 记为$ \hat{\theta}_{B}(x), $即

$ \hat{\theta}_{B}(x) = E(\theta|X = x) = \frac{ \int_{\Theta}\theta f(x|\theta)\mathrm{d}G(\theta)}{f(x)}, x\in(a, b). $

由下一节(3.2)知$ f^{'}(x) = T^{'}(x) \int_{\Theta}\theta f(x|\theta)\mathrm{d}G(\theta)+\frac{h^{'}(x)}{h(x)}f(x) $, 再结合上式可得

$ \begin{eqnarray} \hat{\theta}_{B}(x) = \phi_{B}(x)-\frac{h^{'}(x)}{T^{'}(x)h(x)}, \; \; x\in(a, b) , \end{eqnarray} $

(2.1)

其中$ \phi_{B}(x) = \frac{f^{'}(x)}{T^{'}(x)f(x)} $. 若从总体$ (X, \theta) $抽样得样本$ (X^{*}, \theta^{*}) $, 其中$ \theta^{*} $不可观测, 则$ \hat{\theta}_{B}(X^{*}) $为$ \theta^{*} $的贝叶斯估计. 现先验分布未知, 从而$ f(x) $及$ f^{'}(x) $无法确定, 此时可以考虑先由历史样本$ \{X_{1}, \cdots, X_{n}\} $来估计(2.1)中的$ f(x) $及$ f^{'}(x) $, 由此构造经验贝叶斯决策$ \hat{\theta}_{EB}(x) $, 再将$ X^{*} $代入$ \hat{\theta}_{EB}(x) $中即得到$ \theta^{*} $的经验贝叶斯估计$ \hat{\theta}_{EB}(X^{*}) $. 利用已有样本来估计密度函数的方法有多种, 本文采用递归核估计的方法, 为统一符号分别记$ f(x) $, $ f^{'}(x) $为$ f^{(0)}(x) $, $ f^{(1)}(x) $, 则其递归核估计^[8]分别为

$ \begin{eqnarray} f^{(j)}_{n}(x) = \frac{1}{n}\sum^{n}_{i = 1}\frac{1}{h^{j+1}_{i}}K_{j}\bigg(\frac{x-X_{i}}{h_{i}}\bigg), \; j = 0, 1, \end{eqnarray} $

(2.2)

其中$ K_{j}(\cdot) $为核函数, $ h_{i} $为单调递减趋于0的窗宽. $ f^{(j)}_{n}(x) $具有如下递推性质: $ f^{(j)}_{n}(x) = \frac{n-1}{n}f^{(j)}_{n-1}(x)+\frac{1}{nh^{j+1}_{n}}K_{j}\big(\frac{x-X_{n}}{h_{n}}\big) $, 在运算中可利用此性质减少运算量, 另外, 由于递归核估计的窗宽不同, 可防止估计值过度平滑和过度锐化^[7].

假设此时有$ m $条平行的历史样本$ \{X^{(k)}_{1}, \cdots, X^{(k)}_{n}\}, k = 1, \cdots, m $, 则可考虑先由这$ m $组样本分别利用递归核估计得到$ f^{(j)}_{n, k}(x), k = 1, \cdots, m $, 然后对其求平均值得到$ f^{(j)}(x) $的最终估计$ \bar{f}^{(j)}_{n}(x) = \frac{1}{m}\sum\limits^{m}_{k = 1}f^{(j)}_{n, k}(x) $, 并借此构造如下的经验贝叶斯决策

$ \begin{eqnarray} \tilde{\theta}_{EB}(x, V) = \phi_{EB}(x, V)-\frac{h^{'}(x)}{T^{'}(x)h(x)}, \end{eqnarray} $

(2.3)

其中$ V = \{X^{(k)}_{1}, \cdots, X^{(k)}_{n}, k = 1, \cdots, m\} $, $ \phi_{EB}(x, V) = \big[\frac{\bar{f}^{(1)}_{n}(x)}{T^{'}(x)\bar{f}^{(0)}_{n}(x)}\big]_{n^{\nu}} $, $ [q]_{L} = qI_{\{|q|\leq L\}}(q) $, $ 0<\nu<1 $为一待定常数. 本文会将$ \tilde{\theta}_{EB}(x, V), \; \phi_{EB}(x, V) $简记为$ \tilde{\theta}_{EB}(x), \phi_{EB}(x) $. 若再抽样得$ X^{*} $, 将其代入(2.3)即得$ \theta^{*} $的经验贝叶斯估计$ \tilde{\theta}_{EB}(X^{*}) $.

在平方损失下, 基于单一样本$ X^{*} $的贝叶斯估计风险为$ R(\hat{\theta}_{B}(X^{*})) = E[(\hat{\theta}_{B}(X^{*})-\theta^{*})^{2}] $, 经验贝叶斯估计的风险为$ R(\tilde{\theta}_{EB}(X^{*})) = E[(\tilde{\theta}_{EB}(X^{*})-\theta^{*})^{2}] $. 设$ G(\theta) $属于某先验分布族$ \mathscr{P} $, 在固定$ m $的情况下, 若对分布族$ \mathscr{P} $中每个$ G(\theta) $, 有$ \lim\limits_{n\rightarrow \infty}R(\tilde{\theta}_{EB}(X^{*})) = R(\hat{\theta}_{B}(X^{*})) $, 则称$ \tilde{\theta}_{EB}(X^{*}) $为先验分布族$ \mathscr{P} $下$ \theta^{*} $的渐近最优经验贝叶斯估计. 若更进一步, 对某个$ q>0 $有$ R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*})) = O(n^{-q}) $, 则称此估计在先验分布族$ \mathscr{P} $下的收敛速度为$ O(n^{-q}) $. 下面, 本文将在一定的假设条件下给出当历史样本为强混合时$ \tilde{\theta}_{EB}(X^{*}) $的收敛速度.

记$ C, \; C_{1}, \; C_{2} $为非负常数, $ M_{(x, \delta)} $表示只与$ x, \delta $有关的正数, $ T((a, b)) = \{T(x);x\in(a, b)\} $, $ U(x, \delta) = \{x^{'};|x^{'}-x|<\delta\} $.

所谓强混合序列, 也称$ \alpha $混合序列, 定义如下

定义3.1^[9]  设$ \{\xi_{i}, i = 1, 2, \cdots\} $为随机变量序列, $ \mathscr{F}^{t}_{z} = \sigma(\xi_{i}, \; z\leq i \leq t) $, 若

$ \begin{eqnarray*} \alpha(l)\stackrel{def}{ = }\sup\limits_{w\geq1}\sup\limits_{A\in\mathscr{F}^{w}_{1}, B\in\mathscr{F}^{\infty}_{w+l}}|P(AB)-P(A)P(B)|\rightarrow0.\; \; (l\rightarrow \infty). \end{eqnarray*} $

则称$ \{\xi_{i}, i = 1, 2, \cdots\} $为强混合序列, $ \alpha(\cdot) $称为混合系数.

本文假设条件为

(ⅰ) $ \forall 1\leq k\leq m $, $ \{X^{(k)}_{i}, i\geq1\} $为强混合序列, 其混合系数为$ \alpha^{(k)}(\cdot) $, 且$ \sum^{\infty}_{l = 1}\alpha^{(k)}(l)< \infty $.

(ⅱ) $ (X^{*}, \theta^{*}) $与$ \{(X^{(k)}_{i}, \theta^{(k)}_{i}), i\geq1, k = 1, \cdots, m\} $相互独立.

(ⅲ) 核函数$ K_{j}(x) $可测, 且在$ [-1, 1] $之外为0, 其中$ j = 0, 1 $, 并满足以下条件

$ \int^{1}_{-1}x^{t}K_{j}(x)\mathrm{d}x = \bigg\{^{(-1)^{j}, t = j}_{0, t\neq j, t = 0, 1, \cdots, s-1}, \forall x\!\in R, |K_{j}(x)|\leq C, $

(ⅳ) $ h^{'}(x), T^{'}(x) $在$ (a, b) $上存在, $ T^{'}(x)\neq0 $, 且$ \forall x\in(a, b), \; \exists \delta>0(\delta\; \text{与}\; x\; \text{有关)}, \; \text{使}\; \forall x^{'}\in U(x, \delta) $有

$ \begin{eqnarray} |T(x^{'})-T(x)|\leq M_{(x, \delta)}|x^{'}-x|\qquad T(x)\pm M_{(x, \delta)}\delta\in T((a, b)) \end{eqnarray} $

(3.1)

注1   文献[2, 3, 5, 6]均是在当前样本与历史样本独立的条件下进行研究.

注2   $ \rm(iii) $中的核函数可由多项式函数构造^[1].

注3   若$ T^{'}(x) $在$ (a, b) $上连续, 则$ \forall x\in(a, b) $取$ \delta>0 $足够小使得$ [x-\delta, x+\delta]\subset(a, b) $, 由中值定理可知$ \forall x^{'}\in U(x, \delta) $有$ |T(x^{'})-T(x)|\leq \sup\limits_{ {z\in[x\!-\!\delta\!, x\!+\!\delta]}}\!|T^{'}(z)|\!\times\!|x^{'}-x| $, 因此可取$ M_{(x, \delta)}\! = \!\sup\limits_{ {z\in[x\!-\!\delta\!, x\!+\!\delta]}}\!|T^{'}(z)| $, 同时还有$ \lim\limits_{\delta\rightarrow0^{+}}M_{(x, \delta)}\delta = 0 $, 如果此时$ T((a, b)) $为开集, 则易见(3.1)成立.

引理3.1  设$ f(x|\theta) $是如$ (1.1) $所示的条件分布密度, 当假设$ \rm(iv) $成立时, 对任意先验分布$ G(\theta) $, 边缘分布$ f(x) = \int_{\Theta}f(x|\theta)\mathrm{d}G(\theta) $在$ (a, b) $上具有一阶导数, 且

$ \begin{eqnarray} f^{'}(x) = \frac{h^{'}(x)}{h(x)}f(x)+T^{'}(x)h(x)\int_{\Theta}\theta C(\theta)\exp{\{\theta T(x)\}}\mathrm{d}G(\theta). \end{eqnarray} $

(3.2)

证  由假设(ⅳ)可知$ \forall x\in(a, b), \; \exists\delta>0 $使得$ \forall x^{'}\in U(x, \delta) $有

$ \begin{eqnarray*} \bigg|\frac{\exp{\{\theta(T(x^{'})-T(x))\}-1}}{x^{'}-x}\bigg|& = &\bigg|\sum^{\infty}_{n = 1}\frac{\theta^{n}(T(x^{'})-T(x))^{n}}{n!(x^{'}-x)}\bigg|\\ &\leq&\frac{1}{\delta}\sum^{\infty}_{n = 1}\frac{|\theta\delta M_{(x, \delta)}|^{n}}{n!}\\ &\leq&\frac{1}{\delta}\exp{[|\theta|\delta M_{(x, \delta)}]}. \end{eqnarray*} $

令$ w(x) = \frac{f(x)}{h(x)} = \int_{\Theta}C(\theta)\exp{\{\theta T(x)\}}\mathrm{d}G(\theta) $, 则

$ \begin{eqnarray*} \frac{w(x^{'})-w(x)}{x^{'}-x} = \int_{\Theta}C(\theta)\exp{\{\theta T(x)\}}\bigg(\frac{\exp{\{\theta(T(x^{'})-T(x))\}-1}}{x^{'}-x}\bigg)\mathrm{d}G(\theta). \end{eqnarray*} $

由于$ \forall x^{'}\in U(x, \delta) $结合(3.1)有

$ \begin{eqnarray*} &&\bigg|C(\theta)\exp{\{\theta T(x)\}}\bigg(\frac{\exp{\{\theta(T(x^{'})-T(x))\}-1}}{x^{'}-x}\bigg)\bigg|\\ &\leq&\frac{1}{\delta}C(\theta)\exp{\{\theta T(x)+|\theta|\delta M_{(x, \delta)}\}}\\ &\leq&\frac{1}{\delta}\Big\{C(\theta)\exp{\{\theta T(x_{(1)})\}}+C(\theta)\exp{\{\theta T(x_{(2)})\}}\Big\}. \end{eqnarray*} $

其中$ T(x)+\delta M_{(x, \delta)} = T(x_{(1)}), \; T(x)-\delta M_{(x, \delta)} = T(x_{(2)}) $, $ x_{(1)}, x_{(2)} $与$ x^{'} $无关, 并且

$ \begin{eqnarray*} \int_{\Theta}C(\theta)\exp{\{\theta T(x_{(i)})\}}\mathrm{d}G(\theta) = w(x_{(i)})<+\infty, \; i = 1, 2. \end{eqnarray*} $

所以, 结合控制收敛定理知$ w^{'}(x) $在$ (a, b) $上存在, 且$ w^{'}(x)\! = \!T^{'}(x)\!\int_{\Theta}\!\theta C(\theta)\exp{\![\theta T(x)]}\mathrm{d}G(\theta) $, 由$ f(x) = h(x)w(x) $可知在$ (a, b) $上$ f^{'}(x) $存在, 同时(3.2)成立.

引理3.2  设$ \delta\geq1 $, 则当$ \int_{\Theta}|\theta|^{\delta}\mathrm{d}G(\theta)<+\infty $时, 有$ E[|\hat{\theta}_{B}(X^{*})|^{\delta}]<+\infty $. 如果还有$ E\big[\big|\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{\delta}\big]<\infty $, 则$ E[|\phi_{B}(X^{*})|^{\delta}]<\infty $.

证  利用Jensen不等式知

$ \begin{eqnarray*} &&E[|\hat{\theta}_{B}(X^{*})|^{\delta}] = \int^{b}_{a}|\hat{\theta}_{B}(x^{*})|^{\delta}f(x^{*})\mathrm{d}x^{*} = \int^{b}_{a}|E(\theta^{*}|X = x^{*})|^{\delta}f(x^{*})\mathrm{d}x^{*}\\ &\leq&\int^{b}_{a}E(|\theta^{*}|^{\delta}|X = x^{*})f(x^{*})\mathrm{d}x^{*} = \int^{b}_{a}\int_{\Theta}|\theta^{*}|^{\delta}f(x^{*}|\theta^{*})\mathrm{d}G(\theta^{*})\mathrm{d}x^{*}\\ & = &\int_{\Theta}|\theta^{*}|^{\delta}\mathrm{d}G(\theta^{*})<\infty. \end{eqnarray*} $

再利用$ C_{r} $不等式可得

$ \begin{eqnarray*} E[|\phi_{B}(X^{*})|^{\delta}]& = &E\big[\big|\hat{\theta}_{B}(X^{*})+\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{\delta}\big]\\ &\leq&2^{\delta-1}\Big\{E\big[\big|\hat{\theta}_{B}(X^{*})\big|^{\delta}\big]+E\big[\big|\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{\delta}\big]\Big\}<+\infty. \end{eqnarray*} $

引理3.3^[9]  设$ \{\xi_{i}\}_{i\geq1} $为强混合序列, $ \mathscr{F}^{t}_{z} = \sigma(\xi_{i}, \; z\leq i \leq t), \; \alpha(\cdot)\text{为其混合系数} $. $ \forall w\geq1 $, 若$ \zeta\in\mathscr{F}^{w}_{1}, \eta\in\mathscr{F}^{\infty}_{w+n} $, 且$ |\zeta|\leq C_{1}, |\eta|\leq C_{2} $, 则$ |E(\zeta\eta)-E(\zeta)E(\eta)|\leq 4C_{1}C_{2}\alpha(n) $.

引理3.4^[1]   $ Y, Y^{'} $为随机变量, $ y, y^{'} $为实数, $ L>0, d>0 $, 则

$ \begin{eqnarray*} &&E\Big[\mathrm{min}\Big\{\Big|\frac{Y^{'}}{Y}-\frac{y^{'}}{y}\Big|, L\Big\}^{d}\Big]\leq \mathrm{min}\{L^{d}, A\}. \end{eqnarray*} $

其中$ A = 2^{d+(d-1)^{+}}|y|^{-d}\big\{E|Y^{'}-y^{'}|^{d}+(|\frac{y^{'}}{y}|^{d}+2^{-(d-1)^{+}}L^{d})E|Y-y|^{d}\big\} $.

在接下来我们均令$ \mathscr{F}^{(k)}_{z, t} = \sigma(X^{(k)}_{i}, \; z\leq i \leq t) $, 并指定先验分布族$ \mathscr{P}_{s} $定义如下

$ \mathscr{P}_{s} = \{G(\theta): f^{(s)}(x) 在 \mathbb{R} 上存在, 且 |f^{(s)}(x)|\leq Z_{G}\}, $

其中$ f^{(s)}(x) $表示$ X $的边缘分布$ f(x) $的$ s $阶导数, $ s\geq2 $, $ Z_{G} $为只与$ G(\theta) $有关的非负常数. 注意若$ a\neq-\infty $或$ b\neq+\infty $, 则需要在$ (a, b) $的端点处$ f(x) $的相应阶的右、左导数为零.

引理3.5  设$ G(\theta)\in\mathscr{P}_{s} $, 取$ \rm(2.2) $中的窗宽$ h_{n} = n^{-\frac{1}{2(s+1)}} $, 则$ \forall x\in \mathbb{R}, \; 0<\lambda\leq1, \; k = 1, \cdots, m, \; j = 0, 1 $, 有

$ \begin{eqnarray*} E\big[|f^{(j)}_{n, k}(x)-f^{(j)}(x)|^{2\lambda}\big]\leq D^{(k)}_{G, s, \lambda}n^{-\frac{\lambda(s-1)}{s+1}}, \end{eqnarray*} $

其中$ D^{(k)}_{G, s, \lambda} = \max{\{1, 2^{2\lambda-1}\}}\Big[\big(\frac{4(s+1)Z_{G}C}{(s+3)s!}\big)^{2\lambda}+[C^{2}+32C^{2}\sum\limits^{\infty}_{l = 1}\alpha^{(k)}(l)]^{\lambda}\Big] $.

证  首先, 利用$ C_{r} $不等式有

$ \begin{eqnarray} E\big[|f^{(j)}_{n, k}(x)-f^{(j)}(x)|^{2\lambda}\big]&\leq &\max{\{1, 2^{2\lambda-1}\}}(J_{1}+J_{2}), \end{eqnarray} $

(3.3)

其中$ J_{1} = \big|E[f^{(j)}_{n, k}(x)]-f^{(j)}(x)\big|^{2\lambda} $, $ J_{2} = E\big[|E[f^{(j)}_{n, k}(x)]-f^{(j)}_{n}(x)|^{2\lambda}\big] $, 并且

$ \begin{eqnarray} E[f^{(j)}_{n, k}(x)] = \frac{1}{n}\sum^{n}_{i = 1}\frac{1}{h_{i}^{j+1}}E\Big[K_{j}\Big(\frac{x-X^{(k)}_{i}}{h_{i}}\Big)\Big]. \end{eqnarray} $

(3.4)

注意到$ f(x) $在$ \mathbb{R} $上有定义, 可得

$ \begin{eqnarray} E\Big[K_{j}\Big(\frac{x-X^{(k)}_{i}}{h_{i}}\Big)\Big] = \int^{+\infty}_{-\infty}K_{j}\Big(\frac{x-x^{(k)}_{i}}{h_{i}}\Big)f(x^{(k)}_{i})\mathrm{d}x^{(k)}_{i} = \int^{1}_{-1}K_{j}(t)f(x-th_{i})h_{i}\mathrm{d}t. \end{eqnarray} $

(3.5)

由Taylor公式可得

$ \begin{eqnarray} f(x-h_{i}t) = f^{(0)}(x)-f^{(1)}(x)h_{i}t+\cdots+\frac{f^{(s-1)}(x)(-h_{i}t)^{s-1}}{(s-1)!}+\frac{f^{(s)}(\eta_{t})(-h_{i}t)^{s}}{s!}, \end{eqnarray} $

(3.6)

其中$ \eta_{t} $位于$ x, x-h_{i}t $之间. 将(3.5)(3.6)代入(3.4)再结合假设(ⅲ)得到

$ \begin{eqnarray*} E[f^{(j)}_{n, k}(x)] & = &\frac{1}{n}\sum^{n}_{i = 1}\frac{1}{h^{j}_{i}}\bigg\{(-h_{i})^{j}\Big(\int^{1}_{-1}t^{j}K_{j}(t)\mathrm{d}t\Big)f^{(j)}(x)\!+\!\frac{h^{s}_{i}}{s!}\int^{1}_{-1}K_{j}(t)f^{(s)}(\eta_{t})(-t)^{s}\mathrm{d}t \bigg\}\\ & = &f^{(j)}(x)+\frac{1}{n}\sum^{n}_{i = 1}\frac{1}{s!}\frac{1}{h^{j-s}_{i}}\int^{1}_{-1}K_{j}(t)(-t)^{s}f^{(s)}(\eta_{t})\mathrm{d}t. \end{eqnarray*} $

又因为窗宽$ h_{n}\! = \!n^{-\frac{1}{2(s+1)}} $, 故有$ \frac{1}{n}\sum\limits^{n}_{i = 1}h^{s-j}_{i}\! = \!\frac{1}{n}\sum\limits^{n}_{i = 1}(i^{-\frac{s-1}{2(s+1)}})^{\frac{s-j}{s-1}}\!\leq\!\frac{1}{n}\sum\limits^{n}_{i = 1}i^{-\frac{s-1}{2(s+1)}} \!\leq\!\frac{1}{n} \int^{n}_{0}\!x^{-\frac{s-1}{2(s+1)}}\mathrm{d}x\! = \!\frac{2(s+1)}{s+3}n^{-\frac{s-1}{2(s+1)}} $, 因此可得

$ |E[f^{(j)}_{n, k}(x)]-f^{(j)}(x)|\leq \frac{2Z_{G}C}{s!}\frac{1}{n}\sum\limits^{n}_{i = 1}h^{s-j}_{i}\leq\frac{2(s+1)}{s+3}\frac{2Z_{G}C}{s!}n^{-\frac{s-1}{2(s+1)}}. $

即有

$ \begin{eqnarray} J_{1} = |E[f^{(j)}_{n, k}(x)]-f^{(j)}(x)|^{2\lambda}\leq \Big(\frac{4(s+1)Z_{G}C}{(s+3)s!}\Big)^{2\lambda}n^{-\frac{\lambda(s-1)}{s+1}}. \end{eqnarray} $

(3.7)

接下来考虑$ J_{2} $, 利用Holder不等式有$ J_{2} = E\{|E[f^{(j)}_{n, k}(x)]-f^{(j)}_{n, k}(x)|^{2\lambda}\}\leq \big\{E\big[\big(E[f^{(j)}_{n, k}(x)]-f^{(j)}_{n, k}(x)\big)^{2}\big]\big\}^{\lambda} $, 令

$ \begin{eqnarray} S_{n, k}(x) = \sqrt{n}h^{j+1}_{n}(E[f^{(j)}_{n, k}(x)]-f^{(j)}_{n, k}(x)) = \frac{1}{\sqrt{n}}\sum^{n}_{i = 1}(\frac{h_{n}}{h_{i}})^{j+1}Z_{i, k}(x), \end{eqnarray} $

(3.8)

其中$ Z_{i, k}(x)\! = \!E[K_{j}\!(\frac{x-X^{(k)}_{i}}{h_{i}}\!)]\!-\!K_{j}\!(\frac{x-X^{(k)}_{i}}{h_{i}}\!) $, 而且$ \forall x\!\in\!\mathbb{R}, i\!\in\!\mathbb{N} $有$ |Z_{i, k}(x)|\!\leq\! 2C, E[(Z_{i, k}(x))^{2}]\! = E\big[\big(K_{j}\big(\frac{x-X^{(k)}_{i}}{h_{i}}\big)\big)^{2}\big]\!-\!\big\{E\big[K_{j}\big(\frac{x-X^{(k)}_{i}}{h_{i}}\big)\big]\big\}^{2} \!\leq\!C^{2} $, 即$ Z_{i, k}(x), E[(Z_{i, k}(x))^{2}] $对$ \forall x\in \mathbb{R}\!, i\in \mathbb{N} $一致有界. 又对$ 1\!\leq\! i\!<\!t\!\leq\! n $有$ Z_{i, k}(x)\in\mathscr{F}^{(k)}_{1, i}, Z_{t, k}(x)\in\mathscr{F}^{(k)}_{t, \infty} = \mathscr{F}^{(k)}_{i+(t-i), \infty} $, 并且$ E[Z_{i, k}(x)] = 0 $, 由引理3.3知$ \forall x\in \mathbb{R}, 1\leq i<t\leq n $有$ |E[Z_{i, k}(x)Z_{t, k}(x)]|\leq 16C^{2}\alpha^{(k)}(t-i) $, 从而有

$ \begin{eqnarray*} E[(S_{n, k}(x))^{2}]& = &\frac{1}{n}\bigg[\sum^{n}_{i = 1}\!\big(\frac{h_{n}}{h_{i}}\big)^{2(j+1)}E[(Z_{i, k}(x))^{2}] \!+\!2\!\sum\limits_{1\leq i<t\leq n}\!\big(\frac{h^{2}_{n}}{h_{i}h_{t}}\big)^{j+1}E[Z_{i, k}(x)Z_{t, k}(x)]\bigg]\\ &\leq& \frac{1}{n}\bigg(\sum^{n}_{i = 1}E[(Z_{i, k}(x))^{2}] +2\sum\limits_{1\leq i<t\leq n}\big|E[Z_{i, k}(x)Z_{t, k}(x)]\big|\bigg)\\ &\leq&\frac{1}{n}\Big(nC^{2}+32nC^{2}\sum^{n-1}_{l = 1}\alpha^{(k)}(l)\Big)\\ &\leq&C^{2}+32C^{2}\sum^{\infty}_{l = 1}\alpha^{(k)}(l)<+\infty. \end{eqnarray*} $

由上式与(3.8)可得$ E[(E[f^{(j)}_{n, k}(x)]\!-\!f^{(j)}_{n, k}(x))^{2}]\! = \!\frac{E[(S_{n, k}(x))^{2}]}{nh^{2(j+1)}_{n}}\!\leq\! \big(C^{2}\!+\!32C^{2}\sum^{\infty}_{l = 1}\alpha^{(k)}(l)\big)n^{\frac{j+1}{s+1}-1} $, 即$ J_{2}\leq \big(C^{2}+8C^{2}\sum^{\infty}_{l = 1}\alpha^{(k)}(l)\big)^{\lambda}n^{(\frac{j+1}{s+1}-1)\lambda} $, 结合此式与(3.3)(3.7)可得

$ \begin{eqnarray*} &&E\big[|f^{(j)}_{n, k}(x)-f^{(j)}(x)|^{2\lambda}\big]\\ &\leq&\max{\{1, 2^{2\lambda-1}\}}\bigg[\Big(\frac{4(s+1)Z_{G}C}{(s+3)s!}\Big)^{2\lambda}n^{-\frac{\lambda(s-1)}{s+1}}\!+\!\Big(C^{2}\!+\!32C^{2}\sum^{\infty}_{l = 1}\alpha^{(k)}(l)\Big)^{\lambda}n^{(\frac{j+1}{s+1}-1)\lambda}\bigg]\\ &\leq&\max{\{1, 2^{2\lambda-1}\}}{\bigg[\Big(\frac{4(s+1)Z_{G}C}{(s+3)s!}\Big)^{2\lambda}\!+\!\Big(C^{2}+32C^{2}\sum^{\infty}_{l = 1}\alpha^{(k)}(l)\Big)^{\lambda}\bigg]}n^{-\frac{\lambda(s-1)}{s+1}}. \end{eqnarray*} $

引理3.6  设$ G(\theta)\in\mathscr{P}_{s} $, 取$ \rm(2.2) $中窗宽$ h_{n} = n^{-\frac{1}{2(s+1)}} $, 则$ \forall x\in \mathbb{R}, \; 0\!<\!\lambda\!\leq\!1, j = 0, 1 $, 有

$ \begin{eqnarray*} E\big[|\bar{f}^{(j)}_{n}(x)-f^{(j)}(x)|^{2\lambda}\big]\leq D_{G, m, s, \lambda}n^{-\frac{\lambda(s-1)}{s+1}}, \end{eqnarray*} $

其中$ D_{G, m, s, \lambda} = \frac{\max{\{1, m^{2\lambda-1}\}}}{m^{2\lambda}}\sum\limits^{m}_{k = 1}D^{(k)}_{G, s, \lambda} $.

证  首先易得$ E\big[|\bar{f}^{(j)}_{n}(x)-f^{(j)}(x)|^{2\lambda}\big] \leq E\big[\frac{1}{m^{2\lambda}}\big(\sum\limits^{m}_{k = 1}|f^{(j)}_{n, k}(x)-f^{(j)}(x)|\big)^{2\lambda}\big] $, 再利用$ C_{r} $不等式可得$ \frac{1}{m^{2\lambda}}\big(\sum\limits^{m}_{k = 1}|f^{(j)}_{n, k}(x)-f^{(j)}(x)|\big)^{2\lambda}\leq \frac{\max{\{1, m^{2\lambda-1}\}}}{m^{2\lambda}}\sum\limits^{m}_{k = 1}|f^{(j)}_{n, k}(x)-f^{(j)}(x)|^{2\lambda} $, 结合引理3.5即证.

引理3.7^[10]  设$ \zeta $为定义在概率空间$ (\Omega, \mathscr{F}, P) $上的可积随机变量, $ \mathscr{A}, \mathscr{D} $为$ \mathscr{F} $的两个子$ \sigma $代数, 如果有$ \mathscr{D} $与$ \sigma(\zeta)\vee\mathscr{A} $相独立, 则有$ E[\zeta|\mathscr{D}\vee\mathscr{A}] = E[\zeta|\mathscr{A}] $.

利用类似于文献[1]中引理2.1的证明方法并结合引理3.7可证明如下的引理3.8, 但[1]中样本是独立的.

引理3.8  若$ R(\hat{\theta}_{B}(X^{*}))<+\infty $, $ E\big[(\tilde{\theta}_{EB}(X^{*})-\hat{\theta}_{B}(X^{*}))^{2}\big]<+\infty $, $ E[\theta^{*}]<+\infty $, 则$ R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*})) = E\big[(\tilde{\theta}_{EB}(X^{*})-\hat{\theta}_{B}(X^{*}))^{2}\big] $.

证  首先由$ E\big[(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*}))^{2}\big]<+\infty $, 及

$ \begin{eqnarray*} &&E\Big[\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)^{2} +2\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)\big(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*})\big)\Big]\\ & = &E[(\tilde{\theta}_{EB}(X^{*}, V)-\theta^{*})^{2}]-E[(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*}))^{2}], \end{eqnarray*} $

知$ E\Big[\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)^{2} +2\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)\big(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*})\big)\Big] $存在, 由积分存在可知相关的条件期望存在^[11], 并且有

$ \begin{eqnarray} &&E\Big[\big(\tilde{\theta}_{EB}(X^{*}, V)-\theta^{*}\big)^{2}\Big|X^{*}, V\Big]\\ & = &E\Big[\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)^{2} +2\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)\big(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*})\big)\Big|X^{*}, V\Big]\\ &&+(\tilde{\theta}_{EB}(X^{*}, V)-\hat{\theta}_{B}(X^{*}))^{2}. \end{eqnarray} $

(3.9)

再由$ R(\hat{\theta}_{B}(X)) = E\big[(\hat{\theta}_{B}(X^{*})-\theta^{*})^{2}\big]<+\infty $, $ E[\theta^{*}]<+\infty $得$ E[\hat{\theta}_{B}(X^{*})]<+\infty $, 与前面同理可知相关的条件期望存在, 所以(3.9)等号右边第一项可表示为

$ \begin{eqnarray} E\Big[\big(\hat{\theta}_{B}(X^{*})-\theta^{*}\big)^{2}\Big|X^{*}, V\Big]\!+\!2\Big(\tilde{\theta}_{EB}(X^{*}, V)\!-\!\hat{\theta}_{B}(X^{*})\Big)\Big(\hat{\theta}_{B}(X^{*})\!-\!E\big[\theta^{*}\big|X^{*}, V\big]\Big). \end{eqnarray} $

(3.10)

由假设(ⅱ)结合引理3.7得$ E[\theta^{*}|X^{*}, V]\! = \!E[\theta^{*}|X^{*}]\! = \!\hat{\theta}_{B}(X^{*}) $, 结合(3.10), (3.9)再取期望即证.

引理3.9  设$ A, B, L\in \mathbb{R} $, 若$ |B|\leq \frac{L}{2} $, 则$ |[A]_{L}-B|\leq \mathrm{min}\big\{|A-B|, \frac{3}{2}L\big\} $.

证  当$ L<|A| $时, 有$ |[A]_{L}-B| = |B|<\frac{3}{2}L $, $ |B|\leq\frac{L}{2}<|A|-|B|\leq|A-B| $. 当$ L\geq|A| $时, 有$ |[A]_{L}-B| = |A-B|\leq|A|+|B|\leq\frac{3}{2}L $, 因此引理成立.

下面的定理4.1是本文的主要结果, 其在相应的假设与条件下给出了$ \tilde{\theta}_{EB}(X^{*}) $在先验分布族$ \mathscr{P}_{s} $下的收敛速度.

定理4.1   设先验分布$ G(\theta)\in\mathscr{P}_{s} $, 假设$ \rm(i)–(iv) $成立, 且存在$ \frac{1}{2s}<r<1 $使得$ \int_{\Theta}\theta^{4rs}\mathrm{d}G(\theta)<\!+\infty $, $ E\big[\big|\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{4rs}\big]<\!+\infty $, $ E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\big]<\!+\infty $, $ E\big[(f(X^{*}))^{-2r}\big]<\!+\infty $. 取(2.2)中窗宽$ h_{n} = n^{-\frac{1}{2(s+1)}} $, (2.3)中的$ \nu = \frac{s-1}{4s(s+1)} $, 则有

$ \begin{eqnarray} 0\leq R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*}))\leq Q_{G, m, s, r}n^{-\frac{(s-1)(2rs-1)}{2s(s+1)}} \end{eqnarray} $

(4.1)

其中

$ \begin{align*} &Q_{G, m, s, r}\! = \!A_{G, m, s, r}\!\Big(\!E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\!\big]\!+\!B_{r}E\big[(f(X^{*}))^{-2r}\!\big]\Big)\!+\!9\!\times\!2^{4rs-2}\!E[\phi_{B}(X^{*})^{4rs}]\\ &A_{G, m, s, r} = \big(\frac{3}{2}\big)^{2(1-2r)}2^{2r+(2r-1)^{+}}D_{G, m, s, r}, \; \; \; B_{r} = \frac{1}{2^{2r}}+\big(\frac{3}{2}\big)^{2r}2^{-(2r-1)^{+}} \end{align*} $

即当$ n\rightarrow \infty $时$ R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*})) = O(n^{-\frac{(s-1)(2rs-1)}{2s(s+1)}}) $.

证  由于$ 4rs>2 $, 根据引理3.2可得$ R(\hat{\theta}_{B}(X^{*}))\leq 2\{E[(\hat{\theta}_{B}(X^{*}))^{2}]+E[(\theta^{*})^{2}]\}<+\infty $. 若令$ A_{n} = \{x^{*};|\phi_{B}(x^{*})|\leq \frac{n^{\nu}}{2}\} $, 则当$ x^{*}\in A_{n} $时$ |\phi_{EB}(x^{*})-\phi_{B}(x^{*})|\leq \frac{3}{2}n^{\nu} $, 且

$ \begin{eqnarray} &&E[\phi_{EB}(x^{*})-\phi_{B}(x^{*})]^{2}\\ & = &E[(\phi_{EB}(x^{*})-\phi_{B}(x^{*}))^{2(1-r)}(\phi_{EB}(x^{*})-\phi_{B}(x^{*}))^{2r}]\\ &\leq& (\frac{3}{2}\big)^{2(1-2r)}n^{2(1-r)\nu}E\bigg[\mathrm{min}\Big\{\Big|\frac{\bar{f}^{(1)}_{n}(x^{*})}{T^{'}(x^{*})\bar{f}^{(0)}_{n}(x^{*})}-\frac{f^{(1)}(x^{*})}{T^{'}(x^{*})f^{(0)}(x^{*})}\Big|, \frac{3}{2}n^{\nu}\Big\}^{2r}\bigg] \end{eqnarray} $

(4.2)

$ \begin{eqnarray} &\leq& (\frac{3}{2}\big)^{2(1-2r)}n^{2(1-r)\nu}2^{2r+(2r-1)^{+}}|T^{'}(x^{*})f(x^{*})|^{-2r}\Big\{E\big[|\bar{f}^{(1)}_{n}(x^{*})-f^{(1)}(x^{*})|^{2r}\big]\\ &&+\big[|\phi_{B}(x^{*})|^{2r}+(\frac{3}{2}n^{\nu})^{2r}2^{-(2r-1)^{+}}\big]|T^{'}(x^{*})|^{2r}E\big[|\bar{f}^{(0)}_{n}(x^{*})-f^{(0)}(x^{*})|^{2r}\big]\Big\} \end{eqnarray} $

(4.3)

$ \begin{eqnarray} &\leq& (\frac{3}{2}\big)^{2(1-2r)}n^{2(1-r)\nu}2^{2r+(2r-1)^{+}}|T^{'}(x^{*})f(x^{*})|^{-2r}D_{G, m, s, r}n^{-\frac{r(s-1)}{s+1}}\\ &&\times\Big(1+(\frac{1}{2^{2r}}+(\frac{3}{2})^{2r}2^{-(2r-1)^{+}})|T^{'}(x^{*})|^{2r}n^{2r\nu}\Big) \end{eqnarray} $

(4.4)

$ \begin{eqnarray} &\leq&A_{G, m, s, r}\Big(|T^{'}(x^{*})|^{-2r}+B_{r}\Big)(f(x^{*}))^{-2r}n^{2\nu-\frac{r(s-1)}{s+1}} \end{eqnarray} $

(4.5)

其中(4.2)(4.3)(4.4)分别由引理3.9、引理3.4、引理3.6得到. 再利用(4.5)可得

$ \begin{eqnarray} &&E[(\phi_{EB}(X^{*})-\phi_{B}(X^{*}))^{2}I_{A_{n}}(X^{*})]\\ & = &\int^{b}_{a}E[(\phi_{EB}(x^{*})-\phi_{B}(x^{*}))^{2}]I_{A_{n}}(x^{*})f(x^{*})\mathrm{d}x^{*} \\ &\leq&A_{G, m, s, r}\Big(E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\big]+B_{r}E\big[(f(X^{*}))^{-2r}\big]\Big)n^{2\nu-\frac{r(s-1)}{s+1}}. \end{eqnarray} $

(4.6)

(4.6)处是因为$ X^{*} $与$ V $相独立, 因此有

$ \begin{eqnarray} &&E[(\phi_{EB}(X^{*})\!-\!\phi_{B}(X^{*}))^{2}I_{A_{n}}\!(X^{*})]\\ & = &\int\!(\phi_{EB}(x^{*}\!, \!u)\!-\!\phi_{B}(x^{*}))^{2}I_{A_{n}}\!(x^{*})F_{V}(\mathrm{d}u)\!\times\!F_{X^{*}}(\mathrm{d}x^{*})\\ & = &\int^{b}_{a}E[(\phi_{EB}(x^{*}, V)-\phi_{B}(x^{*}))^{2}]I_{A_{n}}(x^{*})F_{X^{*}}(\mathrm{d}x^{*})\\ & = &\int^{b}_{a}E[(\phi_{EB}(x^{*}, V)-\phi_{B}(x^{*}))^{2}]I_{A_{n}}(x^{*})f(x^{*})\mathrm{d}x^{*}, \end{eqnarray} $

(4.7)

其中$ F_{V}(u), \; F_{X^{*}}(x^{*}) $分别为$ V, X^{*} $的分布函数. 再令$ B_{n} = (a, b)-A_{n} $, 则当$ x^{*}\in B_{n} $时$ |\phi_{EB}(x^{*})-\phi_{B}(x^{*})|<3|\phi_{B}(x^{*})| $, 从而

$ \begin{eqnarray} &&E[\phi_{EB}(X^{*})-\phi_{B}(X^{*}))^{2}I_{B_{n}}(X^{*})]\\ &\leq&9E[(\phi_{B}(X^{*}))^{2}I_{B_{n}}(X^{*})]\\ &\leq&9\big(E[(\phi_{B}(X^{*}))^{4rs}]\big)^{\frac{1}{2rs}}\big(E[I_{B_{n}}(X^{*})]\big)^{\frac{2rs-1}{2rs}}\\ &\leq&9\big(E[(\phi_{B}(X^{*}))^{4rs}]\big)^{\frac{1}{2rs}}\big(E[|2n^{-v}\phi_{B}(X^{*})|^{4rs}]\big)^{\frac{2rs-1}{2rs}}\\ & = & 9\times2^{4rs-2}E[(\phi_{B}(X^{*}))^{4rs}]n^{-2v(2rs-1)} . \end{eqnarray} $

(4.8)

由引理3.2知$ E[(\phi_{B}(X^{*}))^{4rs}]<\infty $, 结合(4.7)(4.8), 代入$ \nu = \frac{s-1}{4s(s+1)} $可得

$ \begin{eqnarray*} E[(\tilde{\theta}_{EB}(X^{*})-\hat{\theta}_{B}(X^{*}))^{2}] = E[(\phi_{EB}(X^{*})-\phi_{B}(X^{*}))^{2}]\leq Q_{G, m, s, r}n^{-\frac{(s-1)(2rs-1)}{2s(s+1)}}. \end{eqnarray*} $

最后由引理3.8即得(4.1).

注4   文献[4]在比本文更强的假设下得到类似结论. 相比文献[4], 本文由递归核估计构造经验贝叶斯估计, 且未对样本的概率密度函数附加更多的限制, 也未对样本平稳性做要求.

固定方差为$ \sigma^{2}_{0} $, 均值为未知参数$ \theta $的正态分布显然满足假设条件(ⅳ). 取先验分布族$ \mathscr{G}_{1} = \{G(\theta);G(\theta)\; \text{均匀分布}\} $, 设$ g(\theta) = \frac{\mathrm{d}G(\theta)}{\mathrm{d}\theta} = \frac{1}{K-T}I_{(T, K)}(\theta) $, 以$ \Psi(x), \psi(x) $分别表示标准正态分布的分布与密度, 此时$ f(x) = \frac{\Psi(\frac{K-x}{\sigma_{0}})-\Psi(\frac{T-x}{\sigma_{0}})}{K-T} $, 易知$ f(x) $存在无限阶有界导数, 因此$ \mathscr{G}_{1}\subset\bigcap_{s\geq2}\mathscr{P}_{s} $. 对任意$ s\geq2 $, 令$ \frac{1}{2s}<r<\frac{1}{2} $, 下面验证定理4.1的条件. 首先$ E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\big] = \sigma^{4r}_{0}E\big[(f(X^{*}))^{-2r}\big] = \frac{\sigma^{4r}_{0}}{(K-T)^{1-2r}}\! \int^{+\infty}_{-\infty}\!\big(\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\!\frac{1}{\sqrt{2\pi}}\exp{\big\{\!-\!\frac{t^{2}}{2}\big\}}\mathrm{d}t\big)^{1-2r}\!\mathrm{d}x $, 记$ I_{1}\! = \! \int^{T}_{-\infty}\!\big(\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\!\frac{1}{\sqrt{2\pi}}\exp{\big\{\!-\!\frac{t^{2}}{2}\big\}}\mathrm{d}t\big)^{1-2r}\!\mathrm{d}x, \; I_{2}\! = \!\int^{K}_{T}\!\big(\!\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\!\frac{1}{\sqrt{2\pi}}\exp{\big\{\!-\!\frac{t^{2}}{2}\big\}}\mathrm{d}t\big)^{1-2r}\!\mathrm{d}x, \; $ $ I_{3}\! = \! \int^{+\infty}_{K}\big(\!\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\!\frac{1}{\sqrt{2\pi}}\exp{\big\{\!-\!\frac{t^{2}}{2}\big\}}\mathrm{d}t\big)^{1-2r}\!\mathrm{d}x $.

因为

$ \begin{eqnarray*} I_{1}&\leq&\int^{T}_{-\infty}\!\Big(\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\frac{1}{\sqrt{2\pi}}\exp{\Big\{\!-\!\frac{(T-x)^{2}}{2\sigma^{2}_{0}}\Big\}}\mathrm{d}t\Big)^{1-2r}\!\mathrm{d}x \leq\sqrt{\frac{2\pi\sigma^{2}_{0}}{1-2r}}\Big(\frac{K-T}{\sigma_{0}\sqrt{2\pi}}\Big)^{1-2r}\!<+\infty\\ I_{2}&\leq&\int^{K}_{T}\Big(\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\frac{1}{\sqrt{2\pi}}\mathrm{d}t\Big)^{1-2r}\mathrm{d}x = \Big(\frac{K-T}{\sqrt{2\pi}\sigma_{0}}\Big)^{1-2r}(K-T)<+\infty\\ I_{3}&\leq&\int^{+\infty}_{K}\!\Big(\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}\frac{1}{\sqrt{2\pi}}\exp{\Big\{\!-\!\frac{(K-x)^{2}}{2\sigma^{2}_{0}}\Big\}}\mathrm{d}t\Big)^{1-2r}\!\mathrm{d}x \leq\sqrt{\frac{2\pi\sigma^{2}_{0}}{1-2r}}\Big(\frac{K-T}{\sigma_{0}\sqrt{2\pi}}\Big)^{1-2r}\!<+\infty, \end{eqnarray*} $

所以$ E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\big]<+\infty, \; E\big[(f(X^{*}))^{-2r}\big]<+\infty $. 又因为$ E\big[\big|\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{4rs}\big] = E[(X^{*})^{4rs}] $, 而

$ \begin{eqnarray*} E[(X^{*})^{4rs}] & = &\frac{1}{K-T}\int^{+\infty}_{-\infty}\int^{\frac{K-x}{\sigma_{0}}}_{\frac{T-x}{\sigma_{0}}}x^{4rs}\psi(y)\mathrm{d}y\mathrm{d}x\nonumber\\ & = &\frac{1}{K-T}\int^{+\infty}_{-\infty}\psi(y)\int^{K-\sigma_{0}y}_{T-\sigma_{0}y}x^{4rs}\mathrm{d}x\mathrm{d}y\nonumber\\ & = &\frac{1}{(K-T)(4rs+1)}E[(K-\sigma_{0}Y)^{4rs+1}-(T-\sigma_{0}Y)^{4rs+1}], \end{eqnarray*} $

其中$ Y\sim N(0, 1) $, 易见$ E\big[\big|\frac{h^{'}(X^{*})}{T^{'}(X^{*})h(X^{*})}\big|^{4rs}\big]<+\infty $. 而$ \int_{\Theta}\theta^{4rs}\mathrm{d}G(\theta)<+\infty $是显然的, 即定理4.1中条件均成立.

再取分布族$ \mathscr{G}_{2} = \{G(\theta);G(\theta)\; \text{为正态分布}\} $. 设$ g(\theta)\! = \!\frac{\mathrm{d}G(\theta)}{\mathrm{d}\theta}\! = \!\frac{1}{\sqrt{2\pi}\sigma}\exp{\{\!-\!\frac{(\theta-\mu)^{2}}{2\sigma^{2}}\}} $, 由于

$ \begin{eqnarray*} f(x)& = &\int^{+\infty}_{-\infty}\!\frac{1}{\sqrt{2\pi}\sigma_{0}}\exp{\Big\{\!-\!\frac{(x-\theta)^{2}}{2\sigma_{0}^{2}}\Big\}}\frac{1}{\sqrt{2\pi}\sigma}\exp{\Big\{\!-\!\frac{(\theta-\mu)^{2}}{2\sigma^{2}}\Big\}}\mathrm{d}\theta\\ & = &\frac{1}{\sqrt{2\pi(\sigma^{2}+\sigma_{0}^{2})}}\exp{\Big\{\!-\!\frac{\sigma^{2}+\sigma_{0}^{2}}{2\sigma^{2}\sigma_{0}^{2}}\Big[\frac{\sigma^{2}x^{2}+\mu^{2}\sigma_{0}^{2}}{\sigma^{2}+\sigma_{0}^{2}}-\Big(\frac{x\sigma^{2}+\mu\sigma_{0}^{2}}{\sigma^{2}+\sigma_{0}^{2}}\Big)^{2}\Big]\Big\}}\\ &&\times\frac{1}{\sqrt{2\pi}\sqrt{\frac{\sigma^{2}\sigma_{0}^{2}}{\sigma^{2}+\sigma_{0}^{2}}}}\int^{+\infty}_{-\infty}\exp{\Big\{\!-\!\frac{\sigma^{2}+\sigma_{0}^{2}}{2\sigma^{2}\sigma_{0}^{2}}\Big(\theta-\frac{x\sigma^{2}+\mu\sigma_{0}^{2}}{\sigma^{2}+\sigma_{0}^{2}}\Big)^{2}\Big\}}\mathrm{d}\theta\\ & = &\frac{1}{\sqrt{2\pi}\sqrt{\sigma^{2}+\sigma_{0}^{2}}}\exp{\Big\{\!-\!\frac{(x-\mu)^{2}}{2(\sigma^{2}+\sigma_{0}^{2})}\Big\}}, \end{eqnarray*} $

因此$ f(x) $为$ N(\mu, \sigma^{2}+\sigma^{2}_{0}) $的概率密度, 易知$ \mathscr{G}_{2}\subset\bigcap_{s\geq2}\mathscr{P}_{s} $. $ \forall s\geq2 $, 令$ \frac{1}{2s}<r<\frac{1}{2} $, 首先由

$ \begin{eqnarray*} E\big[(f(X^{*}))^{-2r}\big] & = &\frac{\sqrt{2\pi(\frac{\sigma^{2}+\sigma^{2}_{0}}{1-2r})}}{[2\pi(\sigma^{2}+\sigma^{2}_{0})]^{\frac{1-2r}{2}}}\int^{+\infty}_{-\infty} \frac{1}{\sqrt{2\pi(\frac{\sigma^{2}+\sigma^{2}_{0}}{1-2r})}}\exp{\bigg\{\!-\!\frac{(x-\mu)^{2}}{2(\frac{\sigma^{2}+\sigma^{2}_{0}}{1-2r})}\bigg\}}\mathrm{d}x\\ & = &\frac{\sqrt{2\pi(\frac{\sigma^{2}+\sigma^{2}_{0}}{1-2r})}}{[2\pi(\sigma^{2}+\sigma^{2}_{0})]^{\frac{1-2r}{2}}}<+\infty, \end{eqnarray*} $

知$ E\big[|T^{'}(X^{*})f(X^{*})|^{-2r}\big] $与$ E\big[(f(X^{*}))^{-2r}\big] $有限, 同时由于正态分布各阶矩存在, 因此其余条件也成立, 所以定理4.1中条件均成立. 注意当$ G(\theta)\in\mathscr{G}_{1}\bigcup\mathscr{G}_{2} $时, 理论上可取足够大的$ s $及$ r\uparrow\frac{1}{2} $和满足条件的核函数使得经验贝叶斯估计的收敛速度趋向$ O(n^{-\frac{1}{2}}) $.

本文参考文献[12]的模拟方式, 在$ \beta $混合样本下($ \beta $混合的定义及与强混合的关系可见[13]), 模拟平均化经验贝叶斯估计的贝叶斯风险差. 首先, 按如下方法可得到一个Gibbs Markov链$ \{(X_{i}, \theta_{i})\}_{1\leq i\leq n} $

(a) 先从先验分布$ g(\theta) = \frac{\mathrm{d}G(\theta)}{\mathrm{d}\theta} $中抽取样本$ \theta_{1} $.

(b) 再从条件分布$ f(x|\theta_{1}) $中取得样本$ X_{1} $.

(c) 再从条件分布$ g(\theta|X_{1}) $中取得样本$ \theta_{2} $.

(d) 重复步骤(a)–(c).

由[14]的推论3.6及[15]的推论13可知$ \{(X_{i}, \theta_{i})\}_{1\leq i\leq n} $为$ \beta $混合的^[12], 从而也是强混合的, 因此可以提出以下模拟方案:

(1) 重复$ m $次抽取Gibbs Markov链的方法抽取$ n $个样本, 得$ \{\!X^{(k)}_{i}, \!1\leq i\leq n\!\}, 1\!\leq\! k\!\leq\! m $.

(2) 利用(1)中$ m $个$ \{X^{(k)}_{i}, 1\leq i\leq n\} $构造$ f^{(j)}_{n, k}(x)(j = 0, 1.\; k = 1, \cdots, m) $, 并求其平均值$ \bar{f}^{(j)}_{n}(x) $, 再将$ \bar{f}^{(j)}_{n}(x) $代入(2.3)中得到平均化的经验贝叶斯决策$ \tilde{\theta}_{EB}(x) $.

(3) 从总体抽取1个样本, 得到$ (X^{*}, \theta^{*}) $, 同时计算$ (\tilde{\theta}_{EB}(X^{*})-\theta^{*})^{2} $与$ (\hat{\theta}_{B}(X^{*})-\theta^{*})^{2} $.

(4) 重复步骤(1)-(3) $ Q $次, 求平均值, 得到$ \hat{R}(\tilde{\theta}_{EB}(X^{*})) = \frac{1}{Q}\sum(\tilde{\theta}_{EB}(X^{*})-\theta^{*})^{2} $与$ \hat{R}(\hat{\theta}_{B}(X^{*})) = \frac{1}{Q}\sum(\hat{\theta}_{B}(X^{*})-\theta^{*})^{2} $, 并将$ \hat{R}(\tilde{\theta}_{EB}(X^{*}))-\hat{R}(\hat{\theta}_{B}(X^{*})) $作为$ R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*})) $的估计值.

取$ f(x|\theta) $为方差$ \sigma^{2}_{0} = 1 $, 均值为参数$ \theta $的正态分布, 考虑先验分布$ g_{1}(\theta) = \frac{1}{10}I_{(-5, 5)}(\theta) $, $ g_{2}(\theta) = \frac{1}{\sqrt{2\pi}}\exp{(-\frac{\theta^{2}}{2})} $, $ g_{3}(\theta) = \frac{1}{\sqrt{2\pi}}\exp{(-\frac{(\theta-1)^{2}}{2})} $. 令$ s = 2 $, 取核函数$ K_{0}(x) = \frac{15}{16}(1-x^{2})^{2}I_{[-1, 1]}(x) $, $ K_{1}(x) = -\frac{15}{4}(1-x^{2})xI_{[-1, 1]}(x) $, 易知$ K_{0}(X) $, $ K_{1}(x) $满足假设(ⅲ). 取$ Q = 200 $, 考虑$ n = 50, \; 100, \; 200 $, $ m = 5, \; 10, \; 20 $, 得出模拟结果见表 1. 由表 1可见, 在固定$ m $的情况下, 随着$ n $增大, 总体而言, $ R(\tilde{\theta}_{EB}(X^{*}))-R(\hat{\theta}_{B}(X^{*})) $的估计值缩小, 这与定理4.1结论相符.